Two lines a and b are perpendicular to each other if the angle between them is 90°.

For intersecting lines we can easily define the perpendicularity. But, if the lines are skew lines then we will take any point on one of the lines and draw a parallel line to the other line. If the angle between these two intersecting lines is 90° then the given skew lines are said to be perpendicular.

Theorem:If one of two parallel lines is perpendicular to a third line, the other is perpendicular too.

Proof:

Let m and b be two parallel lines and m be perpendicular to c (Figure 1.36).

Through any point A, let us draw lines m_{1}and c_{1}so that m_{1}// m and c_{1}// c.

Since m ⊥ c, the angle between m_{1}and c_{1}is 90°.

On the other hand, since m_{1}// m and m // b, we get m_{1}// b.

So the angle between b and c is also 90°.

That means b and c are perpendicular lines

Example 26:By using perpendicularity, prove that if two lines are parallel to the same line they are parallel.

Solution:Let m, n and d be three lines. Let

m // n and d // n. We need to prove that m // d.

Let A be a point on n. We can draw a plane α perpendicular to n at A. So n ⊥ α.

Since m // n and n ⊥ α, m ⊥ α because of the previous theorem.

Since d // n and n ⊥ α then d ⊥ α .

Since m and d are perpendicular to α by using the previous theorem they are parallel.

Check Yourself 8

1.State the followings as true or false.

a.Two lines d and k are parallel to the same plane. If a line l is perpendicular to line d it must be always perpendicular to line k.

b.Two perpendicular lines are given in space. If a line is perpendicular to these lines at their intersection point then this line is perpendicular to the plane, which includes the perpendicular lines.

c.Given that d, k and l are three lines in space. If d and k are perpendicular to line l they must be parallel to each other.

d.Two lines d and k in a plane are perpendicular to the same line l. Another line m is parallel to line l then m is perpendicular to d and k, also.

Answers

1.a) False b) True c) False d) True

B. LINE PERPENDICULAR TO A PLANE

Definition: (line perpendicular to a plane)

A line is said to be perpendicular to a plane if it is perpendicular to every line in this plane.

Let us say that m is a line and α is a plane. If it is given that m ^ α then m is perpendicular to any line in α.

If m ^ α then m intersects α. To prove this statement let us assume that m does not intersect α.

In this case there are two possibilities for m and α :

1.m is in α. Then since it is not perpendicular to a line in α, that is itself, m is not perpendicular to α.

2.m is parallel to α. In this case in α there can be found a line parallel to m. So m can not be perpendicular to α.

In both conditions m is not perpendicular to α.

Therefore, m intersects α.

Definition: (inclined line)

If a line intersects a plane but not perpendicular to the plane it is called an inclined line.

Theorems:

1. If a line is perpendicular to two intersecting lines lying in a plane then it is perpendicular to the plane.

2.Through any given point in space, there can be drawn one and only one plane perpendicular to a given line.

3.If one of two parallel lines is perpendicular to a plane then the other line is also perpendicular to the same plane.

4.Two lines perpendicular to the same plane are parallel.

5.Through a point in space, there can be drawn only one line perpendicular to a given plane.

6.If a line is perpendicular to one of two parallel planes, it is perpendicular to the other.

Proofs:

1.We need to prove that if a line is perpendicular to two intersecting lines in a plane it is perpendicular to any line in this plane.

Let d be a line perpendicular to two lines m and n lying in α. Let A be the intersection point of m and n.

It is obvious that d is perpendicular to every line in α which is parallel to either one of m or n (Figure 1.37).

So we should check for the lines which are not parallel to neither m nor n.

Let x be any line intersecting both m and n.

We have to prove that d is perpendicular to x too.

Let us shift lines d and x so that A is on d and x.

Let c be any line in α intersecting m, n, x at

points C, D, E respectively.

On line d let us take two points B and B' so that BA = B'A.

Then DBAC and DB'AC are congruent, similarly DBAD and DB'AD are congruent (by S.A.S.).

So BD = B'D and BC = B'C.

Then DBDC and DB'DC are congruent (by S.S.S.). That means ∠BDC = ∠B'DC.

Then DBDE and DB'DE are congruent triangles (by S.A.S.).

So BE = B'E and DBAE and DB'AE are congruent (by S.S.S.).

Hence ∠BAE = ∠B'AE = 90°.

So d is perpendicular to x.

Therefore d is perpendicular to any line in α. So d ⊥ α.